For now, let’s look at a gear train with only two gears in it. To be able to find a gear ratio, these gears have to be interacting with each other — in other words, their teeth need to be meshed and one should be turning the other. For example purposes, let’s say that you have one small drive gear (gear 1) turning a larger driven gear (gear 2).
For example purposes, let’s say that the smaller drive gear in our system has 20 teeth.
Let’s say that, in our example, the driven gear has 30 teeth.
In our example, dividing the 30 teeth of the driven gear by the 20 teeth of the drive gear gets us 30/20 = 1. 5. We can also write this as 3/2 or 1. 5 : 1, etc. What this gear ratio means is that the smaller driver gear must turn one and a half times to get the larger driven gear to make one complete turn. This makes sense — since the driven gear is bigger, it will turn more slowly.
Let’s say for example purposes that the two-gear train described above is now driven by a small seven-toothed gear. In this case, the 30-toothed gear remains the driven gear and the 20-toothed gear (which was the driver before) is now an idler gear.
In our example, we would find the gear ratio by dividing the thirty teeth of the driven gear by the seven teeth of our new driver. 30/7 = about 4. 3 (or 4. 3 : 1, etc. ) This means that the driver gear has to turn about 4. 3 times to get the much larger driven gear to turn once.
In our example, the intermediate gear ratios are 20/7 = 2. 9 and 30/20 = 1. 5. Note that neither of these are equal to the gear ratio for the entire train, 4. 3. However, note also that (20/7) × (30/20) = 4. 3. In general, the intermediate gear ratios of a gear train will multiply together to equal the overall gear ratio.
For example, let’s say that in the example gear train above with a seven-toothed driver gear and a 30-toothed driven gear, the drive gear is rotating at 130 RPMs. With this information, we’ll find the speed of the driven gear in the next few steps.
Often, in these sorts of problems, you’ll be solving for S2, though it’s perfectly possible to solve for any of the variables. In our example, plugging in the information we have, we get this: 130 RPMs × 7 = S2 × 30
In our example, we can solve like this: 130 RPMs × 7 = S2 × 30 910 = S2 × 30 910/30 = S2 30. 33 RPMs = S2 In other words, if the drive gear spins at 130 RPMs, the driven gear will spin at 30. 33 RPMs. This makes sense — since the driven gear is much bigger, it will spin much slower.
